The value of K_c = 4.24 at 800K for the reaction

CO(g) + H₂O(g) → CO₂(g) + H₂(g)

Calculate equilibrium concentrations of CO₂, H₂, CO, and H₂O at 800K, if only CO and H₂O are present initially at a concentration of 0.10 M each?

To find the equilibrium concentrations of the gases in the reaction CO(g) + H₂O(g) → CO₂(g) + H₂(g) at 800 K, we can use the equilibrium constant expression and an ICE table (Initial, Change, Equilibrium).

Step 1: Set Up the ICE Table

Initially, we have:

• [CO] = 0.10 M
• [H₂O] = 0.10 M
• [CO₂] = 0 M
• [H₂] = 0 M

Let x be the change in concentration at equilibrium:

• [CO] = 0.10 - x
• [H₂O] = 0.10 - x
• [CO₂] = x
• [H₂] = x

Step 2: Write the Equilibrium Expression

The equilibrium constant expression for the reaction is:

K_c = \(\frac{[CO_2][H_2]}{[CO][H_2O]}\)

Step 3: Substitute the Equilibrium Concentrations

Substituting the equilibrium concentrations into the expression gives:

4.24 = \(\frac{x \cdot x}{(0.10 - x)(0.10 - x)}\)

This simplifies to:

4.24 = \(\frac{x^2}{(0.10 - x)^2}\)

Step 4: Solve for x

Cross-multiplying leads to:

4.24(0.10 - x)^2 = x^2

Expanding and rearranging gives:

4.24(0.01 - 0.20x + x^2) = x^2

0 = (4.24 - 1)x^2 + 0.848x - 0.0424

0 = 3.24x^2 + 0.848x - 0.0424

Step 5: Use the Quadratic Formula

Applying the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) where:

• a = 3.24
• b = 0.848
• c = -0.0424

Calculating the discriminant:

b² - 4ac = (0.848)² - 4(3.24)(-0.0424)

Solving gives two possible values for x. Choose the positive value that makes sense in the context of the problem.

Step 6: Calculate Equilibrium Concentrations

Once you find x, substitute it back to find the equilibrium concentrations:

• [CO] = 0.10 - x
• [H₂O] = 0.10 - x
• [CO₂] = x
• [H₂] = x

These calculations will yield the equilibrium concentrations of CO, H₂O, CO₂, and H₂ at 800 K.